Question

помогите решить Алгебра √32

Answer 1

1)=sinx |от π/6 до5π/6=sin(5π/6)-sin(π/6)=sin(π-π/6) -sin(π/6)=sin(π/6)-sin(π/6)=0;
2)=-cosx |  ot  π/3  do  2π/3=-cos(2π/3) +cosπ/3=-cos(π-π/3) +cosπ/3=
=-(-cosπ/3)+cosπ/3=2*(1/2)=1
3)=((5x^5) /5 +(6x^3) /3 ) | -1 do1=(x^5+2x^3)...=1+2-(-1-2)=3+3=6
4)=4x^4)/4+(6x^2)/2 | ot -2 do 1=(x^4+3x^2)...= 1+3-(16+12)=-24

Answer 2

можно пожалуйста фотографией?)